Plesk: lack of SSD disk space

I'm using Plesk on Ubuntu in the cloud. I'm running out of disk space. Migrating to the next level will double my price.

The cloud company offers a feature to buy a new "volume" that is no more than a directory.

Is there a way to use this volume to unload much of Plesk's content, in order to free the main drive? My backups are stored remotely.

I was told that the answer was no, that I needed to upgrade, but I thought I would ask experts first. If most Plesk data is stored in a single directory, is not it possible to link the two directories?

Geometry – What is the relationship between the geodesic curvature in space and the curvature in the plain?

We are looking for a relationship between the geodesic curvature of a three-dimensional curve in the Euclidean sphere and the curvature of a curve in the Euclidean plain.

We tried to find the relation with the help of a map between the sphere and the plane S², but we could not find the exact relation.

We expect a relationship where kappa = (something) * kappa_ {g}

such as the derivatives of the normal curvature and the geodesic curvature are equal to the same points.

python 3.x – search the longest common substring in the O (n) space

I am working on the class issue to find the longest common substring from 2 strings but I would like to return the longest common substring instead of just the size of the longest common substring.

s1 = 'abcdex', s2 = 'xbcdey' => 'bcde'
s1 = 'abcxyx', s2 = 'xbcydey' => 'bc'

To do this, I use dynamic programming to create a memo table and update each cell requiring O (n * m) space. How can I implement the code so that I can only use space as O (n)?

    def longestCommonSubstring(self, s1, s2):
        :type s: str
        :rtype: str
        m, n = len(s1), len(s2)
        dp = (('') * (n + 1) for _ in range(m + 1))
        max_len = ''
        for i, ic in enumerate(s1):
            for j, jc in enumerate(s2):
                dp(i)(j) = dp(i - 1)(j - 1) + ic if ic == jc else ''
                max_len = dp(i)(j) if len(max_len) < len(dp(i)(j)) else max_len
        return max_len

internal storage – Recover the space occupied by an uninstalled application

I recently uninstalled Economist.
according to Settings -> applications, it took ca. 530 MB of data.
After the uninstallation, this space was not released.

Is it possible to recover this space without installing any utility application?
(I've seen Clean Master be mentioned a few times, I'm not sure of those who mentioned it actually tried, successfully).

Relative, but not providing an answer:

How to recover the space used by the defaul messaging client
(I have reinstalled L & # 39; economist, To see if Settings -> applications signaled the same busy space again, so I could Erase data Of the; but no, it shows only 15 MB of data).

I have uninstalled an application, but I have not released any space

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gm – Is my idea of ​​managing space combat in my environment good?

I'm using a homebrewed system and a challenging science fiction environment (not originally designed for table RPGs but for story writing and as a construction project of people) with a group of friends for a while and they are ready to leave the starting space. station. So, the first space battle is getting closer and closer, but I do not know if my ideas on how to fight in space are useful or not.

The decor is very difficult, science fiction has many millions of years in the future. Therefore, although the technology is very advanced (torch, antimatter, handheld weapons, nanotechnology, genetic engineering, transhumanism), no -fi technologies (FTL readers, teleportation, useful force fields, …) exist . It has worked well so far, but the players have not yet come into contact with really advanced military equipment, such as a spaceship.

The challenging nature of the sci-fi scenery means that space combat is an extremely automated process involving a mind-boggling dance of probability and vector computation, coupled with deception and elaborate bluffs. Evasive maneuvers, firing solutions, and tactical decisions must be calculated, developed, and redone in nanoseconds. This obviously means that human intervention in space combat will be minimal.

So, I thought that if players could not make tactical decisions, win battles for strategic reasons and good preparation would be the way forward. If a normal fight appears to play Fortnite, my space fight would be similar to dota Autochess or Laugh out loudTFT mode: you make strategic decisions and whether their success is positive or not, you must live (or die) with the results. I would describe the strategic options of the players and tell them, as an AI of the spaceship, the probability of success, critical success, failure and critical failure. Thus, the space combat would be very narrative, unlike the usual dice, intensive combat.

It should be noted that in the absence of usable energy shields and spacecraft shielding, the negative consequences would be either the vaporization of the part in a nuclear explosion, overheating and thus the loss of the combat effectiveness of the ship or the loss of a resource or essential component training, ejection mass, ammunition, …) making the victory extremely improbable. This means that fights in space will always involve very important issues, as there is no fantastic way out of bad situations.

My system generally works by adding up all the statistics of a player on a number of dice that he can throw to launch it. Each die is considered a success if he shows more than one of the top 50% of his eyes. (if D6: 1,2,3 == Failed_Roll; 4,5,6 == Succeeded_Roll) For minor decisions, as what type of loot is found, I use a "Destiny_Roll". I run a D20 and ask the player in question if he wants high or low numbers. The result can be a yes / no or a good / bad answer, depending on the situation.

In space combat, they will receive larger dice bonuses due to the equipment of their ship (the addition of a rifle or a dagger shield will bring extra dice) and of a maintenance status (attrition during the battle will clear the dice if a second round occurs). (They will start with an ultra-slow solar sailboat with no armor and only armed with defensive infrared laser networks in the defense, which is rather a highlight for the spacecraft.) They will then discuss their battle plan with them. IA of the spaceship. where they can use their skills and creative suggestions to earn extra dice. In addition, the ship will have limited resources, namely heat capacity, heat sinks, reaction mass, power, armor and bullets whose rate of use depends on the decisions of the players and the result of rolls of dice. These plans are made before the battle or during idle time. (The villains fired a volley of missiles and it will take an hour before the next end arrives so that the strategy can be adjusted.) The short-term adjustments will be managed via "Destiny_Rolls" and will have little to 39; impact.

The decision of the player would be whether the ship should focus on defense or offensive, how many resources it can use, which weapons to prioritize, how much to keep, and so on. Each of these decisions will give a specific bonus to the game, which will not necessarily be the case. have a flat die bonus (sacrifice dice for guaranteed damage if successful, etc.)

What problems could I encounter with this system?

real analysis – $ L ^ p $ space definition and functions defined almost everywhere

Given $ ( Omega, M, mu) $ a measurement space, we define $ L ^ p $ be the quotient of the set of measurable functions $ f $ of $ Omega $ to the actual extended line as $ | f | ^ p $ has a finite integral, with respect to, $ ~ $ the equivalence relationship "almost everywhere equals".

I can not understand why this is a vector space. Take $ f, g $ in $ L ^ p $. Their sum is defined as the class of equivalence of the sum of any two representatives $ h, k $ respectively $ f $ and $ g $. $ h, k $ are functions defined on $ Omega $and are almost everywhere finished, but their sum can be defined almost everywhere. So what are you doing?

I guess we could extend $ h + k $ on the whole field by just putting to $ 0 (or any other value) the sum each time it is not defined. This would always give a well defined sum of equivalence classes. But for the integral? Can I define the integral of a function almost everywhere defined as the integral of the function on the set on which it is defined?

c # – Calculates the position of the object in the screen space – canvas of the camera

I need to calculate the correct position of a 3D object that I display on a RawImage image in my user interface. If I put my UI image in the center of the screen, I can calculate it perfectly. If I move my image around its UI pattern, the results are shifted, depending on which side I move. I've taken some screenshots of what I mean, the orange side is my 3D quad, the white square is just an image debug where my point needs to be calculated.

Correct visualization, image in the center of the screen

Image moved to the left of the screen, erroneous result

The configuration I have is as follows:
– a world camera pointing to a 3D quad
– a canvas of user interface with a dedicated perspective camera (Screen space – Camera)
– a panel in my web of user interface displaying the 3D quad

The code:

var worldPoint = t.Value.MeshRenderer.bounds.min; //t.Value is the 3D quad
var screenPoint = worldCamera.WorldToScreenPoint(worldPoint);
screenPoint.z = (baseCanvas.transform.position - uiCamera.transform.position).magnitude; //baseCanvas is the UI canvas
var pos = uiCamera.ScreenToWorldPoint(screenPoint);
debugger.transform.position = pos; //debugger is just the square image used to see where my calculated point is landing

I've tried several ways, like this:

var screenPoint = worldCamera.WorldToScreenPoint(t.Value.MeshRenderer.bounds.min);
Vector2 localPoint;
RectTransformUtility.ScreenPointToLocalPointInRectangle(rectTransform, screenPoint, uiCamera, out localPoint); //rectTransform is my UI panel
debugger.transform.localPosition = localPoint

But I always get the same result, how can I do the right calculation considering the offset?

MacOS Catalina – Deleting files does not free up space

I know there are more topics on this problem, but I think I have tried all the solutions.

Hi guys,

So I just installed the .1 version of Catalina and I found that I only had about 50 GB left. So I deleted some big unnecessary files and in I transferred others to my external hard drive. But, the disk space did not free up. I have checked all of the internet last night but I can not understand it. Even Mail does not start anymore (because of the low disk space).

That's what I did and some results.

  • Turning off Time Machine (which runs to a machine at the airport time)
  • Reindexd Spotlight
  • Snapshots checked with tmutil. Deleted with them; "tmutil
    listlocalsnapshotdates | grep -e "^ 2" | xargs -n 1 – sudo tmutil
    deletelocalsnapshots "Whenever I check snapshots again, it says
    there are not any
  • I've used several tools to check which files are the largest and where
    the storage is gone. As you can see, this is
    about 400 GB and I have 1TB
  • I made an EHBO disk via the disk utility. The Macintosh HD – Data gave some

I still think there's something with Snapshots, but he says there's more after you delete them.

Another note; All my life, I used "TimeMachineEditor" as a scheduling program for Time Machine. What I turned off. Could this have something to do with the problem? (Do not believe)

May I have help please?

drivers – PCI Linux memory allocation. And if the space allocated by the operating system is incorrect?

I am a noob for PCI and Linux kernel drives.

A few years ago, an article explained how a Linux kernel extracted information from BAR to determine memory requirements and I / O allocation allocation.

In summary, I know, or better still, I think I know, that all 1 are written in a bar. But, not all bits are actually defined by the device. A number of lower bits are not actually written. thus, when the value is read, a number of lower bits are always 0. If 6 bits are always 0, 6 ^ 2 bytes, or 64 bytes, memory is allocated. 8 bits zero. 8 ^ 2 bytes are allocated (256 bytes).

first … did I completely misunderstand it? it seems to make sense to me.

my problem. I have a map whose data manual says that it needs 256 bytes allocated to memory, but when I load it, it only allocates 64 bytes. Thus, all internal registers after the 64th register can not be read.

This is a problem of the bar returning 6 zeros instead of 8 should it be good? Which makes it a manufacturer problem. Or is there a configuration space value that needs to be changed? Or does everything completely depend on the device?

the old message that I mentioned above:
How do the base address registers (BARs) of a PCI card work?